Question

G=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\dfrac{x^2-2x+1}{2}\)

ĐKXĐ:\(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Answer 1

\(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=-\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)